Hi everyone. I need some help if someone might beable to explain this to me. Our

Question

Hi everyone.   I need some help if someone might beable to explain this to me.   Our instructor did not do avery good job explaining a lab very well and now I'm totallylost.   I need some help on some questions I wasgiven.   The data has been changed a little bit (Ichanged the numbers given) so that you won't feel like you're"doing my homework for me".   I was able to answer someof the questions, but these impulse questions are just beyondme!   Please help if anyone can.   I will thankyou in advance. 1.   Find the impulse of the force shown on theforce-time graph below. 2.   An object of mass 2.5 kg is moving in thenegative x direction at a velocity of 2.7 m/s. It experiences theforce shown above for 3 s. What is the final velocity after theobject has experienced the impulse? 3.   A ball was in contact with atable for 34 ms, calculate the average force exerted on theball by the table. Given:   The ball is dropped froma height of 10 cm and its weight is 0.05 kg.   The ballbounces off the table and rises back to the sameheight.     Hint: Firstcalculate the momentum before and after hitting the table. Don'tforget the gravitational force. 4.   A clay ball is dropped with the exact samespecifications as the ball in #3 (ie - weight, height,etc).   Only this time, the clay ball becomes stuck tothe table as it falls onto it.   Assuming the ball iscompletely stuck and doesn't rebound at all during the impact withthe table, calculate the magnitude of the impulse exerted on theclay ball. (Please don't use a sign, which hasn't been definedanyway.) 5.   Suppose that the clay has 2 timesthe mass and is dropped fromthe identical height. Compare the impulse exerted onthe ball by the table to that with the smaller clay ball. 6.   Suppose that the original clay ball is droppedfrom double the height. Compare the impulse exerted on the ball bythe table to that for the smaller height.

Explanation / Answer

1. impulse = force x time interval= 4 x 3 = 12 N.s 2. the force is positive so a = f/m = 4/2.5 m/s2 = 1.6 vf = vo + at= -2.7 + (1.6)(3) = 2.1 m/s 3. the velocity with which the ball impacts the table is givenby v2 = 02 + 2gh v = [(2(9.8)(.1)] = 1.96 = 1.4 after the bounce, the ball has the same speed upwards since itrises to the same height. change in momentum = mv - (-mv) = 2mv = impulse = F(contacttime) F = 2mv/(contact time) = 2(.05)(1.4)/.034 N >>> calculatethis 4. In this case, the change in momentum is mv and so impulse = mv = (.05)(1.4) = .7 kg.m/s 5. The velocity v does not depend on the mass (free fall). so impluse is twice as much as case (4). 6. If the height is 2h, then v = (4gh) >>> compute this impulse = m(4gh) hope this helps!

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